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Alexxx [7]
2 years ago
6

Combine terms/simolify

Mathematics
1 answer:
-Dominant- [34]2 years ago
5 0

Answer:

\dfrac{2m -1}{5}

Step-by-step explanation:

Since we have common denominators inside the parentheses, we can combine the denominators. Then, we get the following expression:

2\huge\text{(}\dfrac{1}{5} m - \dfrac{2}{5} \huge\text{)} + \dfrac{3}{5}

2\huge\text{(}\dfrac{m - 2}{5} \huge\text{)} + \dfrac{3}{5}

Now, we can multiply the expression inside the parentheses by 2.

\huge\text{(}\dfrac{2m - 4}{5} \huge\text{)} + \dfrac{3}{5}

Since the expression inside the parentheses cannot be <u>simplified further</u>, we can open the parentheses and simplify the expression.

\dfrac{2m - 4}{5}  + \dfrac{3}{5}

We can combine the denominators as the denominators are the same.

\dfrac{2m - 4 + 3}{5}

Finally, we can simplify -4 + 3.

\dfrac{2m -1}{5}

Therefore, the simplified expression is (2m - 1)/5.

Learn more about combining like terms: brainly.com/question/13080103

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Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally
rusak2 [61]

Answer:

(a) The probability that the thickness is less than 3.0 mm is 0.119.

(b) The probability that the thickness is more than 7.0 mm is 0.119.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is 0.762.

Step-by-step explanation:

We are given that thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 5.0 millimeters (mm) and a standard deviation of 1.7 mm.

Let X = <u><em>thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village.</em></u>

So, X ~ Normal(\mu=5.0,\sigma^{2} =1.7^{2})

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean thickness = 5.0 mm

           \sigma = standard deviation = 1.7 mm

(a) The probability that the thickness is less than 3.0 mm is given by = P(X < 3.0 mm)

    P(X < 3.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{3.0-5.0}{1.7} ) = P(Z < -1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(b) The probability that the thickness is more than 7.0 mm is given by = P(X > 7.0 mm)

    P(X > 7.0 mm) = P( \frac{X-\mu}{\sigma} > \frac{7.0-5.0}{1.7} ) = P(Z > 1.18) = 1 - P(Z \leq 1.18)

                                                           = 1 - 0.8810 = <u>0.119</u>

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

(c) The probability that the thickness is between 3.0 mm and 7.0 mm is given by = P(3.0 mm < X < 7.0 mm) = P(X < 7.0 mm) - P(X \leq 3.0 mm)

    P(X < 7.0 mm) = P( \frac{X-\mu}{\sigma} < \frac{7.0-5.0}{1.7} ) = P(Z < 1.18) = 0.881

    P(X \leq 3.0 mm) = P( \frac{X-\mu}{\sigma} \leq \frac{3.0-5.0}{1.7} ) = P(Z \leq -1.18) = 1 - P(Z < 1.18)

                                                           = 1 - 0.8810 = 0.119

The above probability is calculated by looking at the value of x = 1.18 in the z table which has an area of 0.881.

Therefore, P(3.0 mm < X < 7.0 mm) = 0.881 - 0.119 = 0.762.

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Answer:

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Step-by-step explanation:

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